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Continuous Functions

The concept of continuous function is basic to much of mathematics. Continuous functions on the real line appear in the first pages of any calculus book, and continuous functions in the plane and in space follow not far behind. More general kinds of continuous functions arise as one goes further in mathematics. In this section, we shall formulate a definition of continuity that will include all these as special cases, and we shall study various properties of continuous functions. Many of these properties are direct generalizations of things you learned about continuous functions in calculus and analysis.

Continuity of a Function

Let and be a topological spaces. A function is said to be continuous if for each open subset of , the set is an open subset of .

Recall that is the set of all points of for which ; it is empty if does not intersect the image set of .

Continuity of a function depends not only upon the function itself, but also on the topologies specified for its domain and range. If we wish to emphasize this fact, we can say that is continuous relative to specific topologies on and .

Let us note that if the topology of the range space is given by a basis , then to prove continuity of it suffices to show that the inverse image of every basis element is open. Then arbitrary open set of can be written as a union of basis elements Then so that is open if each set is open.

If the topology on is given by a subbasis , to prove continuity of it will even suffice to show that the inverse image of each subbasis element is open. The arbitrary basis element for can be written as a finite intersection of subbasis element; it follows from the equation that the inverse image of every basis element is open.

EXAMPLE 1 Let us consider a function like those studied in analysis, a "real-valued function of a real variable," In analysis, one defines continuity of f via the " definition". As one would expect, the definition and ours are equivalent To prove that our definition implies the definition, for instance, we proceed as follows.

Given in , and given , the interval is an open set of the range space . Therefore, is an open set in the domain space . Because contains the point , it contains some basis element about . We choose to be the smaller of the two numbers and . Then if , the point must be in , so that , and , as desired.
Conversely, given in , and given , , if , we have . Let , then , thus is a open set, because is a basis element, as desired.

EXAMPLE 2 In calculus one considers the property of continuity for many kinds of functions. For example, one sudies functions of the following types: Each of them has a notion of continuity defined for it. Our general definition of continuity includes all these as special cases, this fact will be a consequence of general theorems we shall prove concerning continuous functions on product spaces and on metric spaces.

EXAMPLE 3 Let denote the set of real numbers in its usual topology, and let denote the same set in the lower limit topology. Let be the indentity function. for every real number . Then is not a continuous function, the inverse image of the open set of equals itself, which is not open in . On the other hand, the identity function is continuous, because the inverse image of is itself, which is open in .

Theorem 18.1. Let and be topological spaces; let . Then the following are equivalent:

  1. is continuous.

  2. For every subset of , one has .

  3. For every closed set of , the set is closed in .

  4. For each and each neighborhood of , there is a neighborhood of such that .

If the condition in (4) holds for the point of , we say that is continuous at the point .

proof. We show that (1) (2) (3) (1) and that (1) (4) (1).
(1) (2). Assume that is continuous. Let be a subset of . We show that if , then . Let be a neighborhood of . Then is an open set of containing ; it must intersect in some point . THen intersects in the point , so that , as desired.
(2) (3). Let be closed in and let . We wish to prove that is closed in ; we show that . By elementary set theory, we have . Therefore, if , so that . Thus , so that . Thus , so that , as desired.
(3) (1). Let be an open set of . Set . Then Now is a closed set of . Then is closed in by hypothesis, so that is open in , as desired.
(4) (1). Let be an open set of ; let be a point of . Then , so that by hypothesis there is a neighborhood of such that . Then . If follows that can be written as the union of the open sets , so that it is open.
(1) (4). Let and let be a neighborhood of . Then the set is a neighborhood of such that .

Homeomorphisms

Let and be topological spaces; let be a bijection. If both the function and the inverse function are continuous, then is called a homeomorphism.

The condition that be continuous says that for each open set of , the inverse image of under the map is open in . But the inverse image of under the map is the same as the image of under the map . See Figure 18.1. So another way to define a homeomorphism is to say that it is a bijective correspondence such that is open if and only if is open.

世界上听起来最难受的一句话就是:
我原本可以做到的......

最简单的解析几何把题目看错了,第二题求积分,拆 \(\dfrac{1}{(1+sx^{2})(1+x^{2})}\)都拆错了,哎......,偏偏以这种最遗憾的方式结束。

可是没人会为你哀悼,别忘了作业还没写完......

附上答案:初赛试题A卷

一些做题时的details

  1. 面对高代有关线性变换等式的题目,例如给出:

已知两个线性变换, 满足, 证明 的特征值都是 , 且 有公共的特征向量.

这类题目,我们可以考虑将线性变换转换为在对应基的表示矩阵来做,这样一来给出的等式便转换成了矩阵方程,我们可以尝试取迹,看是否能得到与特征值有关的信息,因为我们知道一个矩阵的迹等于其所有特征值的乘积取负。

此外,我们可以尝试对上述矩阵方程左乘或右乘某些矩阵,再取迹,利用迹的性质 , 得到一些新的信息。

  1. 面对可交换的线性变换(给出 ),注意回想可交换的线性变换有哪些隐藏性质!

  2. 注意记清楚函数项级数、反常积分、含参量积分连续、可微、可积的条件。

  3. 对于给出逐点收敛,要证一致收敛的题目,考虑用有限覆盖定理证明。

  4. 对于要放缩的积分,十分注意积分上下限的放缩,比如常见的拆成两部分,一部分趋于 , 一部分趋于一个常数;面对绝对值,更要注意上下限的变换来去掉绝对值;重中之重,面对含三角函数的积分,一定要考虑把积分上下限拆到该三角函数的一个周期中去讨论和放缩!(尤其是面对无穷限的反常积分)

  5. 若对一个连续可微函数,给出某点的导数值,注意考虑 Taylor 展开。

  6. 对于证明多元函数的一些性质的证明,可先考虑一元的情形,用类似思想去证明多元。

  7. 注意基础矩阵 左乘和右乘得到结果的区别:

  8. 面对有关矩阵的题目,我们也要注意将矩阵看作一个线性变换,例如一个可逆矩阵可看作一个同构映射;一个上三角矩阵,可看作不变子空间分解;一个分块对角矩阵,可看作线性空间的直和分解,有根子空间直和分解,循环子空间直和分解,还有不变子空间直和分解,三者分别对应Jordan form、frobenius form 和 直接对角化。

  9. 关于幂零矩阵,注意它与根子空间和jordan form的联系。且记住 这个恒等式,是根子空间分解理论的根源!

  10. 考虑两个矩阵是否相似,先验证不变量是否相同,例如矩阵的迹和行列式,再考虑其特征多项式;考虑两个矩阵是否合同,先考虑是否为是否为对称阵,再考虑将矩阵写为二次型,并化为规范型,计算其正负惯性指数。

  11. 对于给出同时含的原函数和导函数的不等式条件,尝试考虑构造新的函数,利用已知不等式证明构造的函数具有单调性,进而解决问题。对于构造的思路,可以考虑乘指数函数 ,注意指数可以多变 (可以是一个抽象函数,也可以是一个具体函数), 但还是具体情况具体分析,这类问题十分多变。

  12. 面对一些复杂积分,可以尝试用复变函数理论去计算。

  13. 对于给出数列递推式的问题,多去考虑取对数,利用整体法将一个复杂的同项看作一个新的数列,还有就是根据题目形式利用下面两个恒等式估计通项

Fighting!

Closed Sets and Limit Points

Closed Sets

A subset of a topological space is said to be closed if the set is open.

EXAMPLE 1 The subset of is closed because its complement is open. Similarly, is closed, because its complement is open. These facts justify our use of the terms "closed interval" and "closed ray" The subset of is neither open nor closed.

EXAMPLE 2 In the plane , the set is closed, because its complement is the union of the two sets each of which is a product of open sets of and is, therefore, open in

EXAMPLE 3 In the finite complement topology on a set , the closed sets consist of itself and all finite subsets of

finite complement topology

The finite complement topology (or cofinite topology) on a set is defined by specifying the open sets in the topology:

1.The empty set and the set itself are open.

2.Any subset is open if and only if its complement is finite.

This topology is particularly useful when studying properties of convergence and compactness, as it ensures that every sequence in has a convergent subsequence.

EXAMPLE 4 In the discrete topology on the set , every set is open, it follows that every set is closed as well.

EXAMPLE 5 Consider the following subset of the real line: in the subspace topology. In this space, the set is open, since it is the intersection of the open set of with . Similarly, is open as a subset of ; it is even open as a subset of . Since and are complements in of each other, we conclude that both and are closed as subsets of .


Let's talk about the properties of the closed sets.

Theorem 17.1. Let be a topological space. Then the following condition hold:

  1. and are closed.

  2. Arbitrary intersections of closed sets are closed.

  3. Finite unions of closed sets are closed.

proof. (1) and are closed because they are the complements of the open sets and , respectively.
(2) Given a collection of closed sets , we apply DeMorgan's law, Since the sets are open by definition, the right side of this equation represents an arbitrary union of open sets, and is thus open. Therefore, is closed.
(3) Similarly, if is closed for , consider the equation The set on the right side of this equation is a finite intersection of open sets and is therefore open, Hence is closed.

We should notice the conceptions of "closed set" is relative.

Now when dealing with subspaces, one needs to be careful in using the term "closed set." If is a subspace of , we say that a set is closed in if is a subset of and if is closed in the subspace topology of (that is, if is open in )

We have the following theorem:

Theorem 17.2. Let be a subspace of . Then a set is closed in if and only if it equals the intersection of a closed set of with .

proof. Assume that , where is closed in . (See Figure 17.1) Then is open in , so that is open in , by the definition of the subspace topology. But . Hence is open in , so that is closed in . Conversely, assume that is closed in . (See Figure 17.1) Then is open in , so that by definition it equals the intersection of an open set of with . The set is closed in , and , so that equals the intersection of a closed set of with , as desired.

Figure 17.1

Theorem 17.3. Let be a subspace of . If is closed in and is closed in then is closed in .

proof. If is closed in , then equals the intersection of a closed set of with , by the theorem 17.2, so we have . And is also closed in , thus is closed in .

Closure and Interior of a Set

Given a subset of a topological space , the interior of is defined as the union of all open sets contained in .

The closure of is defined as the intersection of all closed sets containing

The interior of is denoted by Int and the closure of is denoted by . Obviously Int is an open set and is a closed set; furthermore, If is open , ; while if is closed, .

We shall not make much use of the intenor of a set, but the closure of a set will be quite important.


NOTE:

When dealing with a topological space and a subspace , one needs to exercise care in taking closures of sets If is a subset of , the closure of in and the closure of in will in general be different In such a situation.

We reserve the notation to stand for the closure of in . The closure of in can be expressed in terms of .

Theorem 17.4. Let be a subspace of , let be a subset of , let denote the closure of in . Then the closure of in equals .

proof. Let denote the closure of in . The set is closed in , so is closed in in Theorem 17.2. Since contains , and since by definition equals the intersection of all closed subsets of containing , we must have
On the other hand, we know that is closed in . Hence by Theorem 17.2, for some set closed in . Then is a closed set containing ; because is the intersection of all such closed sets, we conclude that . Then .

Since the collection of all closed sets in , like the collection of all open sets, is usually much too big to work with. Another way of describing the closure of a set, useful because it involves only a basis for the topology of , is given in the following theorem.

Theorem 17.5. Let be a subset of the topological space .

  1. Then if and only if every open set containing intersects .

  2. Supposing the topology of is given by a basis, then if and only if every basis element containing intersects .

proof. Consider the statement in (a). It is a statement of the form . Let us transform each implication to its contrapositive, thereby obtaining the logically equivalent statement (not ) (not ). Written out, it is the following. In this form, our theorem is easy to prove. If is not in , the set is an open set containing that does not intersect , as desired. Conversely, if there exists an open set containing which does not intersect , then is a closed set containing . By definition of the closure , the set must contain , therefore, cannot be in . Statement (b) follows readily if every open set containing intersects , so does every basis element containing , because is an open set. Conversely, if every basis element containing intersects , so does every open set containing , bacause contains a basis element that contains .

Mathematicians often use some special terminology here. They shorten the statement " is an open set containing " to the phrase Using this terminology, one can write the first half of the preceding theorem as follows:

If is a subset of the topological space , then if and only if every neighborhood of intersects .

EXAMPLE 6 Let be the real line . If , then , for every neighbrhood of intersects , while every point outside has a neighborhood disjoint from Similar arguments apply to the following subsets of .

If , then . If , then . If is the set of rational numbers, then . If is the set of positive integers, then . If is the set of positive reals, then the closure of is the set .

EXAMPLE 7 Consider the subspace of the real line . The set is subset of , its closure in is the set , and its closure in is the set

Limit Point

There is yet another way of describing the closure of a set, a way that involves the important concept of limit point, which we consider now.

If is a subset of the topological space and if is a point of , we say that is a limit point (or "cluster point," or "point of accumulation") of if every neighborhood of intersects in some point other than itself. Said differently, is a limit point of if it belongs to the closure of . The point may lie in or not; for this definition it does not matter.

EXAMPLE 8 Consider the real line . If , then the point is a limit point of and so is the point . In fact, every point of the interval is a limit point of , but no other point of is a limit point of .

If , then is the only limit point of . Every other point of has a neighborhood that either does not intersect at all, or it intersects only in the point itself. If , then the limit points of are the points of the interval .

If is the set of rational numbers, every point of is a limit point of . If is the set of positive integers, no point of is a limit point of . If is the set of positive reals, then every point of is a limit point of .

Theorem 17.6. Let be a subset of the topological space , let be the set of all limit points of . Then

proof. If in , every neighborhood of intersects (in a point different from x). Therefore, by Theorem 17.5, belong to . Hence . Since by definition , it follows that .
To demonstrate the reverse inclusion, we let be a point of and show that . If happens to lie in , it is trivial that ; suppose that does not lie in . Since , we know that every neighborhood of intersects ; because , the set must intersect in a point different from . Then , so that , as desired.

Corollary 17.7 A subset of a topological space is closed if and only if it contains all its limit points.

proof. The set is closed , and the latter holds .

Hausdorff Spaces

One's expenence with open and closed sets and limit points in the real line and the plane can be misleading when one considers more general topological spaces.

For example, in the spaces and , each one-point set is closed. This fact is easily proved; every point different from has a neighborhood not intersecting , so that is its own closure.
But this fact is not true for arbitrary topological spaces. Consider the topology on the three-point set indicated in Figure 17.3. In this space, the one-point set is not closed, for its complement () is not open.
Figure 17.3

And, in topological space , a subset of can have many different limit points.

Similarly, one's experience with the properties of convergent sequences in and can be misleading when one deals with more general topological spaces. In an arbitrary topological space, one says that a sequence of points of the space to the point of provided that, corresponding to each neighborhood of , there is a positive integer such that for all . In and , a sequence cannot converge to more than one point, but in an arbitrary space, it can. In the space indicated in Figure 17.3, for example, the sequence defined by setting for all converges not only to the point , but also to the point and to the point !

Topologies in which one-point sets are not closed, or in which sequences can converge to more than one point, are considered by many mathematicians to be somewhat strange. They are not really very interesting, for they seldom occur in other branches of mathematics.

Therefore, one often imposes an additional condition that will rule out examples like this one, bringing the class of spaces under consideration closer to those to which one's geometric intuition applies. The condition was suggested by the mathematician Felix Hausdorff, so mathematicians have come to call it by his name.

Definition. A topological space is called a Hausdorff space if for each pair of distinct points of , there exists neighborhoods , and of and , respectively, that are disjoint.

Theorem 17.8. Every finite point set in a Hausdorff space is closed.

proof. It suffices to show that every one-point set is closed. If is a point of different from , then and have disjoint neighborhoods and , respectively. Since does not intersect , the point cannot belong to the closure of set . As a result, the closure of the set is itself, so that is is closed.

The condition that finite point sets be closed is in fact weaker than the Hausdorff condition.

For example, the real line in the finite complement topology is note a Hausdorff space, but it is a space in which finite point sets are closed.

The condition that finite point sets be closed has been given a name of its own: it is called the axiom

Theorem 17.9. Let be a space satisfying the axiom; let be a subset of . Then the point is a limit point of if and only if every neighborhood of contains infinitely many points of .

proof. If every neighborhood of intersects in infinitely many points, it certainly intersects in some point other than itself, so that is a limit point of .
Conversely, suppose is a limit point of , and suppose some neighborhood of intersects in only finitely many points. Then also intersects in finitely many points; let be the points of . The set is an open set of , since the finite point set is closed; then is a neighborhood of x that intersects the set not at all. This contradicts the assumption that is a limit point of .

One reason for our lack of interest in the axiom is the fact that many of the interesting theorems of topology require not just that axiom, but the full strength of the Hausdorff axiom. Furthermore, most of the spaces that are important to mathematicians are Hausdorif spaces. The following two theorems give some substance to these remarks.

Theorem 17.10. If is a Hausdorff space, then a sequence of points of converges to at most one point of .

proof. Suppose that is a sequence of points of that converges to . If , let and be disjoint neighborhoods of and , respectively. Since contains for all but finitely many values of , the set cannot. Therefore, cannot converge to .

If the sequence of points of the Hausdorff space converges to the point of , we often write , and we say that is the limit of the sequnce .

Theorem 17.11.

(1). Every simply ordered set is a Hausdorff space in the order topology.

(2). The product of two Hausdor-if spaces is a Hausdorff space.

(3). A subspace of a Hausdorff space is a Hausdorff space.

proof.
(1). Given two elements of the simply orderd set , and . We can find an interval be the neighborhood of , and be the neighborhood of such that . Thus, , is a Hausdorff space.
(2). Suppose that and are the two Hausdorff spaces, and is the product topology. Given , then . We can find a neighborhood of , and of such that , we also can find a neighborhood of , and of such that . Hence , and . So is a Hausdorff space.
(3). Suppose is a Hausdorff space, . Obviously, is also a Hausdorff space.

Order Topology

If is a simply ordered set, there is a standard topology for , defined using the order relation. It is called the order topology; in this section, we consider it and study some of its properties.

Suppose that is set having simple order relation . Given elements and of such that , there are four subsets of that are called the intervals determined by and . They are the following:

The notation used here is familiar to you already in the case where X is the real line, but these are intervals in an arbitrary ordered set. A set of the first type is called an open interval in X, a set of the last type is called a closed interval in X, and sets of the second and third types are called half-open intervals

Definition. Let be a set with a simple order relation; assume has more than one element. Let be the collection of all sets of the following types:

  1. All open intervals in .

  2. All intervals of the form , where is the smallest element (if any) of .

  3. All intervals of the form , where is the largest element (if any) of .

The collection is a basis for a topology on , which is called the order topology.

if has no smallest element, there are no sets of type (2), and if X has no largest element, there are no sets of type (3)

Note: The elements of the order topology are not the intervals, we should notice the definition of the interval!!!

We need to check that satisfies the requirements for a basis.

First, note that every element of lies in at least one element of : The smallest element (if any) lies in all sets of type (2), the largest element (if any) lies in all sets of type (3), and every other element lies in a set of type (1).
Second, note that the intersection of any two sets of the preceding types is again a set of one of these types, or is empty.

EXAMPLE 1 The standard topology on , as defined in the preceding section is just the order topology derived from the usual order on

EXAMPLE 2 Consider the set in the dictionary order, we shall denote the general element of by , to avoid difficulty with notation. The set has neither a largest nor a smallest element, so the order topology on , so the order topology on has as basis the collection of all open intervals of the form for , and for , and . These two types of intervals are indicated below.The subcollection consisting of only intervals of the second type is also a basis for the order topology on , it is easy to check, just think a moment.

EXAMPLE 3 The positive integers form an orederd set with a smallest element. The order topology on is discrete topology, for every one-point set is open If , then the one-point set is a basis element; and if , the one-point set is a basis element.

EXAMPLE 4 The set in the dictionary order is another example of an ordered set with a smallest element. Denoting by and by , we can represent by The order topology on is not the discrete topology. Most one-point sets are open, but there is an exception --- the one-point set .Any open set containing b1 must contain a basis element about (by definition), and any basis element containing contains points of the sequence.


Rays

Definition. If is an ordered set, and is an element of , there are four subsets of that are called the rays determined by They are the following: Sets of the first two types are called open rays, and sets of the last two types are called closed rays

The use of the term "open" suggests that open rays in X are open sets in the order topology. And so they are.

Consider, for example, the ray . If has a largest element , then equals the basis element . If has no largest element, then equals the union of all basis elements of the form , for . In either case, is open. A similar argument applies to the ray .

The open rays, in fact, form a subbasis for the order topology on , as we now show.

Because the open rays are open in the order topology, the topology they generate is contained in the order topology.
On the other hand, every basis element for the order topology equals a finite intersection of open rays; the interval equals the intersection of and , while and , if they exist, are themselves open rays. Hence the topology generated by the open rays contains the order topology.

The Product Topology on X × Y

If and are topological spaces, there is a standard way of defining a topology on the cartesian product .

Definition. Let and be topological spaces. The product topology on is the topology having as a basis the collection of all sets of the form , where is an open subset of and is an open subset of .

Let's check that is a basis.

The first condition is trivial, since is itself a basis element. The second condition is almost as easy, since the intersection of any two basis elements and is another basis element. For and the latter set is a basis element because and are open in and , respectively. See the figure below, it can help you understand it intutively.

Note: The collection is not a topology on . The union of the two rectangles pictured in the figure above, for instance, is not a product of two sets, so it is cannot belong to ; however, it is open in .


We may ask what's the basis of the topology of

Theorem 15.1. If is a basis for the topology of and is a basis for the topology of , then the collection is a basis for the topology of .

proof. We apply Lemma 13.2. Given an open set of and a point of , by definition of the product topology there is a basis element such that . Because and are bases for and , respectively, we can choose an element such that , and an element such that . Then . Thus the collection meets the criterion of Lemma 13.2, so is a basis for .

EXAMPLE 1 We have a standard topology on the order topology. The product of this topology with itself is called the standard topology on . It has as basis the collection of all products of open sets of , but the theorem just proved tells us that the much smaller collection of all products of open intervals in will also serve as a basis for the topology of . Each such set can be pictured as the interior of a rectangle in


It is sometimes useful to express the product topology in terms of a subbasis. To do this, we first define certain functions called projections.

Definition. Let be defined by the equation let be defined by the equation The maps and are called the projections of onto its first and second factors, respectively.

We use word "onto" because and are surjective.

If is an open subset of , then the set is precisely the set , which is open in . Similarly, if is open in , then , which is also open in . The intersection of these two sets is the set , as indicated in Figure below. This fact leads to the following theorem:

Theorem 15.2. The collection is a subbasis for the product topology on .

proof. Let denote the product topology on . let be the topology generated by . Because every element of belongs to , so do the arbitrary unions of finite intersections of elements of . Thus . On the other hand, every basis element for the topology of is a finite intersection of elements of , since Therefore, , so that , hence , as desired.

The subspace Topology

Definition. Let be a topological space with topology . If is a subset of , the collection is a topology on , called the subspace topology. With this topology, is called a subspace of ; its open sets consists of all intersections of open sets of with .

It is easy to see that is a topology.

It contains and because where and are elements of .
The fact that it is closed under finite intersections and arbitrary unions follows from the equations

Lemma 16.1. If is a basis for the topology of then the collection is a basis for the subspace topology on .


proof. Given open in and given , we can choose an element such that . Then . It follows from Lemma 13.2 that is a basis for the subspace topology on .

When dealing with a space and a subspace , one needs to be careful when one uses the term "open set". Does one mean an element of the topology of or an element of the topology of ?

We make the following definition If is a subspace of , we say that a set U is open in (or open relative to ) if it belongs to the topology of ; this implies in particular that it is a subset of . We say that U is open in if it belongs to the topology of

There is a special situation in which every set open in is also open in .

Lemma 16.2. Let be a subspace of . If is open in and is open in , then is open in .

proof. Since is open in , for some set open . Since and are both open in , so is .

Now let us explore the relation between the subspace topology and the order and product topologies.

Theorem 16.3. If is a subspace of and is a subspace of , then the product topology on is the same as the topology inherits as a subspace of .

proof.The set is the general basis element for , where is open in and is open in . Therefore, is the general basis element for the subspace topology on . Now Since and are the general open sets fot the subspace topologies on the and , respectively, the set is the general basis element for the product topology on .
Thus, we know that the bases for the subspace topology on and for the product topology on are the same. Hence the topologies are the same.

Now let be an ordered set in the order topology, and let be a subset of . The order relation on , when restricted to , makes into an ordered set.

However, the resulting order topology on need not be the same as the topology that inherits as a subspace of . We give one example where the subspace and order topologies on agree, and two examples where they do not.

EXAMPLE 1 Consider the subset of the real line , in the subspace topology. The subspace topology has as basis all sets of the form , where in an open interval in . Such a set is of one the following types. By definition, each of these sets is open in . But sets of the second and third types are not open in the larger space .

Note that these sets form a basis for the order topology on Y. Thus, we see that in the case of the set , its subspace topology (as a subspace of ) and its order topology are the same.

EXAMPLE 2 Let be the subset of . In the subspace topology on the the one-point set is open, because it is the intersection of the open set with . But in the order topology on , the set is not open. Any basis element for the order topology on that contains is of the form for some , such a set necessarily contains points of less than .

EXAMPLE 3 Let . The dictionary order on is just the restriction to of the dictionary order on the plane . However, the dictionary order topology on is not the same as the subspace topology on obtained from the dictionary order topology on ..

For example, the set is open in in the subspace topology, but not in the order topology, because we choose , any basis element for the order topology on that contains is of the form such a set is necessarily not be contained in

The set in the dictionary order topology will be called the order square, and denoted by


Convex

Given an ordered set , let us say that a subset of is convex in if for each pair of points of , the entire interval (a, b) of points of lies in . Note that intervals and rays in are convex in .

Theorem 16.4 Let be an orderd set in the order topology; let be a subset of that is convex in . Then the order topology on is the same as the topology inherits as a subspace of .

proof. Consider the ray in . What is its intersection with ? If , then this is an open ray of the orderd set . If , then is either a lower boud on or an upper bound on , since is convex. In the former case, the set equals all of ; in the latter case, is is empty.
A similar work remark show that the intersection of the ray with is either an open ray of , or itself, or empty. Since the sets and form a subbasis for the subspace topology on , and since each is open in the order topology, the order topology contains the subspace topology.
To prove the reverse, note that any open ray of Y equals the intersection of an open ray of with , so it is open in the subspace topology on . Since the open rays of are a subbasis for the order topology on , this topology is contained in the subspace topology, hence they are the same.

Comments: In this proof, we need to be familiar with the definition of rays and its properties (such that it is the subbasis for the order topology)

To avoid ambiguity, let us agree that whenever is an ordered set in the order topology and is a subset of , we shall assume that is given the subspace topology unless we specifically state otherwise. If is convex in , this is the same as the order topology on , otherwise, it may not be.

The exterior measure

Definition. If is any subset of , the exterior measure of is where the infimum is taken over all countable coverings by closed cubes. In general, the exterior measure is always non-negative butt could be infinite, hence we have

Info Note

  1. It is important to note that it would not suffice to allow finite sumsin the definition of

  2. One can, however, replace the coverings by cubes, with covering by rectangles; or with coverings by balls. That the former alternative yields the same exterior measure is quite direct.

EXAMPLE 1. The exterior measure of a point is zero. The exterior measure of the empty set is also zero.

This is clear that a point is a cube with volume zero, and which covers itself, so does the empty set.

EXAMPLE 2. The exterior measure of a closed cube is equal to its volume.

Suppose is a closed cube in . Since covers itself, we have .Therefore, it suffices to prove the reverse inequality.
We consider an abitraty covering by cubes, and by the definition of the exterior measure, it suffices to prove that For a fixed we choose for each an open cube which contains , and such that . From the open covering of the compact set , we may select a finite subcovering, which after possibly renumbering the rectangles, we may write as . Taking the closure of the cubes , we have . Consequently, Since is arbitrary, we find the inequality before we need to prove holds; thus , as desired.

EXAMPLE 3. If is an open cube, the result still holds.

Since is covered by its closure , and , we immediately see that , and obviously by the definition of exterior measure.

EXAMPLE 4. The exterior measure of a rectangle is equal to its volume.

Arguing as in Example 2, we see that . To obtain the reverse inequality, consider a grid in formed by cubes of sides length . Then, if consists of the (finite) collection of all cubes entirely contained in , and the (finite) collection of all cubes that intersect the complement of , we first note that . Also, a simple argument yields Moreover, there are cubes in , and these cubes have volume , so that . Hence and letting tend to infinity yields , as desired.

EXAMPLE 5. The exterior measure of is infinite.

This follows from the fact that any covering of is also a covering of any cube , hence . Since can have arbitrarily large volume, we must have

EXAMPLE 6. The Cantor set has exterior measure .

From the construction of , we know that , where each is a disjoint union of closed intervals, each of length . Consequently, for all , hence .


Observation 1 (Monotonicity) If , then .

This follows that any coverings of by a countable collection of cubes is also a covering of .

In particular, monotonicity implies that every bounded subset of has finite exterior measure.

Observation 2 (Countable sub-additivity) If , then .

First, we may assume that each , for otherwise the inequality clearly holds. For any , the definition of exterior measure yields for each a covering by closed cubes with Then, is a covering of by closed cubes, and therefore Since this holds true for every , the second observation is proved.

Observation 3 If , then , where the infimum is taken over all open sets containing .

By monotonicity, it is clear that the inequality holds. For the reverse inequality, let and choose cubes such that , with Let denote an open cube containing , and such taht . Then is open, and by Observation 2 Hence , as was shown.

Observation 4 If , and , then

By Observation 2, we already know that , so it is suffices to prove the reverse inequality. To this end, we first select such that . Next, we choose a covering by closed cubes, with . We may, after subdividing the cubes , assume that each has a diameter less than . In this case, each can intersect at most one of the two sets or . If we denote by and the sets of those indices for which intersects and , respectively, then is empty, and we have Therefore

Observation 5 If a set is the countable union of almost disjoint cubes , then

Let denote a cube strictly contained in such that , where is a arbitrary but fixed. Then, for every , the cubes are disjoint, hence at a finite distance from one another, and repeated applications of Observation 4 imply Since , we conclude that for every integer , When the tend to infinity, we deduce for every , hence , as desired.

Info Note

This last property shows that if a set can be decomposed into almost disjoint cubes, its exterior measure equals the sum of the volumes of the cubes. Moreover, this also yields a proof that the sum is independent of the decomposition

Measureable sets and the Lebesgue measure

Definition. A subset of is Lebesgue measurable or simply measureable, if for any there exists an open set with and

If is measurable, we define its Lebesgue measure (or measure) by Clearly, the Lebesgue measure inherits all the features contained in Observations 1 - 5 of the exterior measure.


Let's study the properties of the Lebesgue measure.

Property 1 Every open set in is measurable.

Property 2 If , then is measurable. In particular, if is a subset of a set of exterior measure , then is measurable.

By Observation 3 of the exterior measure, for every there exists an open set with and . Since , monotonicity implies , as desired.

Property 3 A countable union of measurable sets is measurable.

Suppose , where each is measurable. Given , we may choose for each an open set with and . Then the union is open, , nd , so monotonicity and sub-additivity of the exterior measure imply

Property 4 Closed sets are measurable

First, we observe that it suffices to prove that compact sets are measurable. Indeed, any closed set can be written as the union of compact sets, say , where denotes the closed ball of radius centered at the origin; then Property 3 applies.
So suppose is compact (so that in particular ), and let . Since is closed, the difference is open, and by Theorem 1.4 we may write this difference as a countable union of almost disjoint cubes FOr a fixed , the finite union is compact; therefore (we isolate this little fact in a lemma below). Since , Observation 1, 4, and 5 of the exterior measure imply Hence , and this also holds in the limit as tends to infinity. Invoking the sub-additivity property of the exterior measure finally yields as desired.

Theorem 1.4. Every open subset of , , can be written as a countable union of almost disjoint closed cubes.

Lemma 3.1. If is closed, is compact, and these sets are disjoint, then .

proof. Since is closed, for each point , there exists so that . Since covers , and is compact, we may find a subcover, which we denote by . If we let , then we must have . Indeed, if and , then for some we have , and by construction . Therefore and the lemma is proved.

Compact set: A bounded set is compact if it is also closed. Compact sets enjoy the Heine-Borel covering property.

Property 5 The complement of a measurable set is measurable.

If E is measurable, then for every positive integer we may choose an open set with and . The complement is closed, hence is measurable, which implies that the union is also measurable by Property 3. Now we simply note that , and such that for all . Therefore, , and is measurable by Property 2. Therefore is measurable since it is the union of two measurable sets, namely and .

Property 6 A countable intersection of measurable sets is measurable.

This follows from Properties 3 and 5, since

Warning

we have proved that the collection of measurable sets is closed under countable unions and intersections. We emphasize, however, that the operations of uncountable unions or intersections are not permissible when dealing with measurable sets!

Theorem 3.2. If are disjoint measurable sets, and , then

proof. First, we assume further that each is bounded. Then, for each , by applying the definition of measurability to , we can choose a closed subset or with . For each fixed , the set are compact and disjoint, so that . Since , we must have Letting tend to infinity, since was arbitrary we find that Since the reverse inequality always holds(by sub-aditivity in Observation 2), this concludes the proof when each is bounded.
In the general case, we select any sequence of cubes that increases to , in the sense that for all and . We then let and for . If we define measurable sets by . then The union above is disjoint and every is bounded. Moreover , and this union is also disjoint. Putting these facts together, and using what has already been proved, we obtain as claimed.


We make two definitions to state succinctly some further consequences.

If is a countable collection of subsets of that increases to in the sense that for all , and , then we write

Similarly, if decreases to in the sense that for all k, and , we write .

Corollary 3.3 Suppose are measurable subsets of .

  1. If , then .

  2. If and for some k, then

proof. For the first part, let , , and in general for . By their construction, the sets are measurable, disjoint, and . Hence and since we get the desired limit.
>For the second part, we may clearly assume that . Let for each , so that is a disjoint union of measurable sets. As a result, we find that Hence, since , we see that , and the proof is complete.

Info Note

The reader should note that the second conclusion may fail without the assumption that for some . This is shown by the simple example when , for all .


What follows provides an important geometric and analytic insight into the nature of measurable sets, in terms of their relation to open and closed sets.

an arbitrary measurable set can be well approximated by the open sets that contain it, and alternatively, by the closed sets it contains.

Theorem 3.4 Suppose is a measurable subset of . Then, for every :

  1. There exists an open set with and .

  2. There exists a closed set with and .

  3. If is finite, there exists a compact set with and .

  4. If is finite, there exists a finite union of closed cubes such that The notation stands for the symmetric difference between the sets and , defined by , which consists of those points that belong to only one of the two sets or .

proof. Part(i) is just the definition of the measurability.

For the second part, we know that is measurable, so there exists an open set with and . If we let , then is closed, , and . Hence as desired.

Comments: If we meet the closed set, we can consider its complement, because its complement is open set.

For (iii), we first pick a closed set so that and . For each , we let denote the ball centered at the origin of radius and define compact sets . Then is a sequence of measurable that decreases to , and since , we conclude that all large one has .

For (iv), choose a family of closed cubes so that Since , the series converges and exists such that . If , then

Invariance properties of Lebesgue measure

Translation-invariance: If is a measurable set and , then the set is alse measurable, and .

To prove the measurability of under the assumption that is measurable, we note if is open, , and , then is open, , and

Relative dilation-invariance: Suppose , and denote by the set . We can then assert that is measurable whenever is, and .

Reflection-invariant: Whenever is measurable, so is and .

- algebras and Borel sets

Definition. A - algebra of sets is a collection of subsets of that is closed under countable unions, countable intersections, and complements.

NOTE: We should compare the definition of - algebra with the topology. In the definition of the topology, we only need to guarantee that finite intersections is closed, but the union can be arbitrary, instead of countable. This is a difference of the two important definition.

The collection of all subsets of is of course a - algebra.

Another - algebra, which plays a vital role in analysis, is the Borel - algebra in , denoted by , which by definition is the smallest - algebra that contains all open sets. Elements of this - algebra are called Borel sets.

We should discuss the existence and uniqueness of the Borel -algebra.

The definition of the Borel -algebra will be meaningful once we have defined the term “smallest,” and shown that such a -algebra exists and is unique. The term "smallest" means that if is any -algebra that contains all open sets , then necessarily . Since we observe that any intersection (not necessarily countable) of -algebras is again a -algebra, we may define as the intersection of all -algebras that contain the open sets. This shows the existence and uniqueness of the Borel -algebra.

Since open sets are measurable, we conclude that the Borel -algebra is contained in the -algebra of measurable sets. Naturally, we may ask if this inclusion is strict.

do there exist Lebesgue measurable sets which are not Borel sets? The answer is “yes.”

Starting with the open and closed sets, which are the simplest Borel sets. Next in order would come countable intersections of open sets; such sets are called sets. Alternatively, one could consider their complements, the countable union of closed sets, called the sets.

Corollary 3.5 A subset of is measurable

  1. if and only if differs from a by a set of measure zero.

  2. if and only if differs from an by a set of measure zero.

proof. Clearly is measurable whenever it satisfies either (i) or (ii), since the , , and sets of measure zero are measurable.
Conversely, if is measurable, then for each integer we may select an open set that contains , and such that . Then is a that contains , and for all . Therefore for all ; hence has exterior measure zero, and is therefore measurable.
For the second implication, we simpy apply part (ii) of Theorem 3.4 with , and take the union of the resulting closed sets.

Construction of a non-measurable set

Are all subsets of measurable? In this section, we answer this question when by constructing a subset of which is not measurable. This justifies the conclusion that a satisfactory theory of measure cannot encompass all subsets of

The construction of a non-measurable set uses the axiom of choice, and rests on a simple equivalence relation among real numbers in .

We write whenever is rational, and note that this is an equivalence relation since the follwing properties hold:

  • for every

  • if , then

  • if and , then .

Two equivalence classed either are disjoint or coincide, and is the disjoint union of all equivalence classed, which we write as Now we construct the set by choosing exactly one element from each , and setting .

Theorem 3.6 The set is not measurable.

proof. The proof is by contradiction, so we assume that is measurable. Let be an enumeration of all the rationals in , and consider the translates We claim that the sets are disjoint, and To see why these sets are disjoint, suppose that the intersection is non-empty. Then there exist rationals and and with ; hence Consequently and is rational; hence , which contradicts the fact that contains only one representative of each equivalence class.
The second inclusion is straightforward since each is contained in by construction. Finally, if , then for some , and therefore for some . Hence , and the first inclusion holds. Now we may conclude the proof of the theorem. If were measurable, then so would be for all , and since the union is disjoint, the inclusions in (4) yield Since is a tanslate of , we must have for all . Consequently, This is the desired contradiction, since neither nor is possible.

Axiom of choice

That the construction of the set is possible is based on the following general proposition.

  • Suppose is a set and is a collection of non-empty subsets of . (The indexing set of 's is not assumed to be countable.) Then there is a function (a "choice function") such that for all . In this general form this assertion is known as the axiom of choice.

The initial realization of the importance of this axiom was in its use to prove a famous assertion of Cantor, the well-ordering principle. This proposition (sometimes referred to as "transfinite induction") can be formulated as follows.

A set is linearly oredered if there is a binary relation such that:
(a) x for all .
If are distinct, then either or (but not both).
If and , then .

We say that a set can be well-orederd if can be linearly oredered in such a way that every non-empty subset has a smallest element in that ordering (that is, an element such that for any other ).

EXAMPLE. A simple example of a well-ordered set is , the positive integers with their usual ordering. The fact that is well-ordered is an essential part of the usual (finite) induction principle. More generally, the well-ordering principle states:

  • Any set can be well-ordered.

It is in fact nearly obvious that the well-ordering principle implies the axiom of choice: if we well-order , we can choose to be the smallest element in , and in this way we have constructed the required choice function. It is also true, but not as easy to show, that the converse implication holds, namely that the axiom of choice implies the well-ordering principle.

The fact that the axiom of choice and the well-ordering principle are equivalent is a consequence of the following considerations.

One begins by defining a partial ordering on a set to be a binary relation on the set that satisfies:

  1. x for all .

  2. If and , then .

  3. If and , then .

If in addition or whenever , then is a linear ordering of .

The axiom of choice and the well-ordering principle are then logically equivalent to the Hausdorff maximal principle:

Every non-empty partially ordered set has a (non-empty) maximal linearly ordered subset.

In other words, if is patially ordered by , then contains a non-empty subset which is linearly ordered by and such that if is contained in a set also linearly ordered by , then .

An application of the Hausdorff maximal principle to the collection of all well-orderings of subsets of E implies the well-ordering principle for E. However, the proof that the axiom of choice implies the Hausdorff maximal principle is more complicated.

Measurable functions

With the notion of measurable sets in hand, we now turn our attention to the objects that lie at the heart of integration theory: measurable functions.

The starting point is the notion of a characteristic function of a set , which is defined by The next step is to pass to the functions that are the building blocks of integration theory. For the Riemann integral it is in effect the class of step functions, with each given as a finite sum where each is a rectangle, and the are constants.

However, for the Lebesgue integral we need a more general notion, as we shall see in the next chapter. A simple function is a finite sum where each is a measurable set of finite measure, and the are constants.

Definition and basic properties

We begin by considering only real-valued functions on , which we allow to take on the infinite values and , so that belongs to the extended real numbers We shall say that is finite-valued if for all . In the theory that follows, and the many applications of it, we shall almost always find ourselves in situations where a function takes on infinite values on at most a set of measure zero.

A function defined on a measurable subset of is measurable, if for all , the set is measurable. To sumplify our notation, we shall often denote the set simply by whenever no confusion is possible.

Basis for a Topology

Definiton. If is a set, a basis for a topology on is a collection of subsets of (called basis elements) such that

  1. For each , there is at least one basis element containing
  2. If , then there is a basis element containing such that

If satisfies these two conditions, then we define the topology of generated by as follows: A subset of is said to be open in (that is, to be an element of ) if for each there is a basis such that and

Note: Every basis element is itself an element of

Example 1. Let be the collection of all circular regions (interiors of circles) in the plane. Then satisfies both conditions for a basis. The second condition is illustrated in Figure 13 1. In the topology generated by , a subset of the plane is open if every x in lies in some circular region contained in

Example 2. Let be the collection of all rectangular regions (intenors of rectangles) in the plane, where the rectangles have sides parallel to the coordinate axes Then satisfies both conditions for a basis. The second condition is illustrated in Figure 13. 2; in this case, the condition is trivial, because the intersection of any two basis elements is itself a basis element (or empty) As we shall see later, the basis generates the same topology on the plane as the basis given in the preceding example.

Example 3. If is any set, the collection of all one-point subsets of is a basis for the discrete topology on

We need to check that the collection is indeed a topology on .

Let us check now that the collection generated by the basis is a topology on .

If is the empty set, it satisfies the defining condition of openness vacuously. Now let us take an indexed family of elements of and show that Given , there is an index such that . Since is open, there is an basis element such that , then , so that is open, by definition.
Now let us take two elements and of and show that . Given , choose a basis element containing such that , and containing such that . Hence , and by the second condition for a basis, there is a basis element such that . Then , so , by definition.
Finally, we show by induction that any finite intersection of elements of is in . This fact is trivial for ; we suppose it true for and prove it for . Now By hypothesis, ; by the result just proved the intersection of and also belong to .
Thus we have checked that collection of open sets generated by a basis is, in fact, a topology.


Another way of describing the topology generated by a basis is given in the following lemma:

Lemma 13.1. Let be a set,let be a basis for a topology on . Then equals the collection of all unions of elements of

proof. we need to prove any subset of satisfies the equation First given a collection of elements of ,there are also elements of . Because is a topology, their union is in . Conversely, given . Choose each , there exists an element of such that , thus , and it is obvious that , Thus, .

This lemma states that every open set in can be expressed as a union of basis elements. This expression for is not, however, unique. Thus the use of the term "basis" in topology differs drastically from its use in linear algebra, where the equation expressing a given vector as a linear combination of basis vectors is unique.

We have described in two different ways how to go from a basis to the topology it generates. Sometimes we need to go in the reverse direction, from a topology to a basis generating it. Here is one way of obtaining a basis for a given topology; we shall use it frequently.

Lemma 13.2. Let be a topological space. Suppose that is a collection of open sets of such that for each open set of and each in , there is an element of such that . Then is a basis for the topology of .

proof. We must show that is a basis, thus we need to check the two conditions of definition of basis, given , since is itself an open set, there is by hypothesis an element of such that . To check the second condition, let , where and are elements of . Since and are open, so is , then there exists by hypohesis an element such that .
Let be a collection of open sets of ; we must show that the topology generated by equals the topology . First, note that if and if , then there is by hypohesis an element of such that . If follows that , by definition. Conversely, if , then by the preceding lemma. Since each element of belongs to and is a topology, also belong to .


We can use the bases to determine whether one topology is finer than another.

Lemma 13.3. Let and be bases for the two topologies and on , respectively.Then the following states are equivalent:

  1. is finer than .

  2. For each and each basis element containing , there is a basis element such that .

proof. (2) (1). Given a element of , we wish to show that . Let , since generate , there is a basis element such that . Condition (2) tell us that there exist an element such that , then , so , by definition.
(1) (2). Given a element and , then , and by condition (1), we have , thus . Since is generated by , there is an element such that .

Example 4. We can see that the collection of all circular regions in the plane generates the same topology as the collection of all rectangular regions.


We now define three topologies on the real line , all of which are of interest.

Definition. If is a collection of all open intervals in the real line, the topology generated by is called the standard topology on the real line. WhWhenever we consider , we shall suppose it is given this topology unless we specifically state otherwise. If is the collection of all half-open intervals of the form where , the topology generated by is called the lower limit topology on . When is given the lower limit topology, we denote it by 𝕝. Finally, let denote the set of all numbers of the form , for , and let be the collection of all open intervals, along with the set of the form . The topology generated by will be called - topology on . When is given by this topology, we denote it by

It is easy to see that all three of these collections are bases; in each case, the intersection of two basis elements is either another basis element or is empty. The relation between these topologies is the following.

Lemma 13.4. The topologies of and are strictly finer than the standard topology on , but are not comparable with another.

proof. Let be the topologies of , respectively. Given a basis element for and a point , there exists a basis element such that , thus is finer than by the Lemma 13.3.
A similar argument applies to . Given a basis element for and a point , there exist a basis element that contains . On the other hand, given a basis element for and point of , there is no open interval contains and lies in .


Subbasis

Definition. A subbasis for a topology on is a collection of subsets of whose union equals . The topology generated by the subbasis is defined to be the collection of all unions of finite intersections of elements of .

We must of course check that is a topology. By the definition, we know that the elements of are all the finite intersections of elements of , and we define the collection of all finite intersections of elements of is . By the lemma 13.1, it will suffice to show that is a basis, and is a topology. So we just need to check the two conditions of basis for . Given , it belongs to an element of and hence to an element of ; this is the first condition for a basis. To check the second condition, let be two elements of , then the is also a finite intersection of elements of , so it belongs to .Obviously, .

Above all, is a topology.

hexo渲染latex编写的数学公式真是一言难尽啊,给我整了好久,之前说什么latex的"|"会被nunjucks模板误解,导致编译错误,然后我先去掉出错的代码,运行

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接着,直接重新把之前出错的代码再粘贴回去,刷新服务器,我发现原来出错的代码可以正确显示公式,现在我再运行
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又发现它又不报错了,无语......

对于有界区间的划分,其范数定义为.现设上函数满足Lipschitz条件,即存在常数,使得对任何,成立.定义 存在,则称曲线可求长.记等分.证明:

(1).存在 (2).曲线 可求长

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